I was testing the uniform convergence of the series $\sum\limits_{n = 1}^{\infty} \dfrac{2^n x^{2^n - 1}}{1 + x^{2^n}}$ on $\left| x \right| \leq \dfrac{1}{2}$.
I wanted to invoke the Wiestrauss M test. For that, I tried
$$\left| \dfrac{2^n x^{2^n - 1}}{1 + x^{2^n}} \right| < \dfrac{2^n \left| x^{2^n - 1} \right|}{\left| x^{2^n} \right|} = \dfrac{2^n}{\left| x \right|}$$
But, I am stuck here since I cannot get the required inequality in terms of $\left| f_n \left( x \right) \right| \leq M_n$ and $\sum\limits_{n = 1}^{\infty} M_n$ is convergent so that the series $\sum\limits_{n = 1}^{\infty} f_n \left( x \right)$ is uniformly convergent.
If $f_n(x)=\dfrac{2^nx^{2^n-1}}{1+x^{2^n}}$, then $f_n$ is an odd function and it is strictly increasing. So, the maximum of $\lvert f_n\rvert$ is attained at $\dfrac12$, and that maximum is $\dfrac{2^{n-2^n+1}}{1+2^{-2^n}}$. It is not hard to see that the series $\displaystyle\sum_{n=1}^\infty\frac{2^{n-2^n+1}}{1+2^{-2^n}}$ converges, and therefore your series converges uniformly, by the Weierstrass $M$-test.